Question

Q4 Find the points of discontinuity, if any, for the function
f(x) = { x2-x-2 / x +1, x not equal to -1
-1 ,X= -1 }

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Given function is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{\dfrac{ {x}^{2} - x - 2}{x + 1} \: \: if \: x \: \ne \: - \: 1} \\ &\sf{ - 1 \: \: when \: x \: = \: - \: 1} \end{cases}\end{gathered}\end{gathered}$

Now, we have to check the continuity of the function at x = - 1.

We know,

➢ A function f(x) is said to be Continuous at x = a, iff

$\rm :\longmapsto\:f(a) = \displaystyle\lim_{x \to a}f(x)$

Thus,

Consider,

$\rm :\longmapsto\:f( - 1) = - 1$

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to \: - \: 1} \: f(x)$

$\rm \: = \: \: \displaystyle\lim_{x \to - 1} \: \dfrac{ {x}^{2} - x - 2 }{x + 1}$

$\rm \: = \: \: \dfrac{ {( - 1)}^{2} - ( - 1) - 2}{ - 1 + 1}$

$\rm \: = \: \: \dfrac{1 + 1 - 2}{0}$

$\rm \: = \: \: \dfrac{0}{0}$

$\rm \: = \: \: which \: is \: meaningless$

So,

$\rm \: = \: \: \displaystyle\lim_{x \to - 1} \: \dfrac{ {x}^{2} - x - 2 }{x + 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to - 1} \: \dfrac{ {x}^{2} - 2x + x - 2 }{x + 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to - 1} \: \dfrac{x(x - 2) + 1( x - 2 )}{x + 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to - 1} \: \dfrac{(x - 2)( x + 1)}{x + 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to - 1}(x - 2)$

$\rm \: = \: \: - 1 - 2$

$\rm \: = \: \: - 3$

$\bf :\longmapsto\:\displaystyle\lim_{x \to \: - \: 1} \: f(x) = - 3$

Hence,

$\bf :\longmapsto\:\displaystyle\lim_{x \to \: - \: 1} \: f(x) \: \ne \: f( - 1)$

Hence,

$\bf :\longmapsto\:f(x) \: is \: not \: continuous \: at \: x = \: - \: 1$

Therefore,

$\bf :\longmapsto\:Only \: one \: point \: of \: discontinuity.$